Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $z = \dfrac{n - 6}{n^2 - 8n + 12} \times \dfrac{-4n^2 + 8n}{-8n - 40} $
Answer: First factor the quadratic. $z = \dfrac{n - 6}{(n - 2)(n - 6)} \times \dfrac{-4n^2 + 8n}{-8n - 40} $ Then factor out any other terms. $z = \dfrac{n - 6}{(n - 2)(n - 6)} \times \dfrac{-4n(n - 2)}{-8(n + 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (n - 6) \times -4n(n - 2) } { (n - 2)(n - 6) \times -8(n + 5) } $ $z = \dfrac{ -4n(n - 6)(n - 2)}{ -8(n - 2)(n - 6)(n + 5)} $ Notice that $(n - 6)$ and $(n - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -4n(n - 6)\cancel{(n - 2)}}{ -8\cancel{(n - 2)}(n - 6)(n + 5)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $z = \dfrac{ -4n\cancel{(n - 6)}\cancel{(n - 2)}}{ -8\cancel{(n - 2)}\cancel{(n - 6)}(n + 5)} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $z = \dfrac{-4n}{-8(n + 5)} $ $z = \dfrac{n}{2(n + 5)} ; \space n \neq 2 ; \space n \neq 6 $